arXiv:1412.3066v3 [math.CO] 25 Jun 2015

Anti-Ramsey Theory on Complete Bipartite Graphs Stephan Cho∗

Jay Cummings† Claire Sonneborn§

Colin Defant‡

June 26, 2015

Abstract We consider quadruples of positive integers (a, b, m, n) with a ≤ b and m ≤ n such that any proper edge-coloring of the complete bipartite graph Km,n contains a rainbow Ka,b subgraph. We show that any such quadruple with m ≥ a and n > (a2 − a + 1)(b − 1) satisfies this property and find an infinite sequence where this bound is sharp. We also define and compute some new anti-Ramsey numbers.

1

Introduction

An edge-colored graph is said to be rainbow if no two edges have the same color. Similarly, an edge-coloring of a graph is said to be a proper coloring if no two adjacent edges have the same color. A typical anti-Ramsey problem concerns properly edgecoloring complete graphs Kn in order to forbid or guarantee the existence of certain rainbow subgraphs. However, proper edge-colorings of complete bipartite graphs have received considerably less attention. It is our goal to prove a few basic results about proper edge-colorings of complete bipartite graphs in an anti-Ramsey-theoretic setting. Namely, we will investigate quadruples of positive integers (a, b, m, n) with ∗

Dept. of Mathematics, UC Berkeley, USA [email protected]. Supported by NSF grant no. 1262930. † Dept. of Mathematics, UC San Diego, USA [email protected]. ‡ Dept. of Mathematics, University of Florida, USA [email protected]. Supported by NSF grant no. 1262930. § University of Dayton, USA [email protected],. Supported by NSF grant no. 1262930.

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a ≤ b and m ≤ n such that every proper edge-coloring of Km,n contains at least one rainbow Ka,b subgraph. We denote this property by Km,n →R Ka,b . We will make use of the following canonical correspondence between properly edge-colored complete bipartite graphs and latin rectangles. Let G be a properly edge-colored copy of Km,n . Let A be the set of m nonadjacent vertices in G, and let B be the set of n nonadjacent vertices in G. Then we may construct an m × n latin rectangle R so that each row of R corresponds to a vertex in A and each column in R corresponds to a vertex in B. Furthermore, every symbol in R corresponds to the color assigned to the edge connecting the vertices that correspond to the row and column in which that symbol is placed. The fact that no two adjacent edges in G have the same color corresponds to the fact that no symbol appears more than once in any row or column of R. We will always convene to let an m × n latin rectangle be one with m rows and n columns. An x × y subrectangle of a latin rectangle R is the intersection of x rows and y columns of R. Finally, we will use the word “rainbow” to describe any subrectangle whose symbols are all distinct.

2

General bounds

Theorem 2.1. Let a, b, m, n be positive integers such that a ≤ b, a ≤ m ≤ n, and n > (a2 − a + 1)(b − 1). Every properly edge-colored Km,n contains a rainbow Ka,b subgraph. That is, Km,n →R Ka,b . Proof. It suffices to show that any properly edge-colored Ka,n contains a rainbow Ka,b subgraph. Suppose we have a proper edge-coloring of Ka,n , and let R be the corresponding a × n latin rectangle. Because R is latin, we may choose any column of R to obtain a rainbow a × 1 subrectangle of R. Now, suppose that we have managed to find a rainbow a × t subrectangle of R for some t ∈ {1, 2, . . . , b − 1}, and call this subrectangle T . Each of the at distinct symbols in T may appear at most a − 1 times outside of T because it can appear no more than once in each row of R. Furthermore, because there are n − t columns of R outside of T and n − t > (a2 − a + 1)(b − 1) − t ≥ (a2 − a + 1)t − t = at(a − 1), we see that there is some column of R containing none of the at symbols that appear in T . We may annex this additional column to T to form a rainbow a × (t + 1) subrectangle of R. By induction, we see that we may construct a rainbow a × b subrectangle of R, so the proof is complete. It is natural to ask how good the bound n > (a2 − a + 1)(b − 1) used in Theorem 2.1 is. In other words, if n ≤ (a2 − a + 1)(b − 1), can we always find a proper 2

edge-coloring of Km,n that forbids the appearance of any rainbow Ka,b subgraphs? Theorem 2.3 makes progress toward answering this question. First, we need a couple of preliminary results. Definition 2.1. Let P be the set of all possible orders of a finite projective plane (we convene to let 1 ∈ P). We will make use of a theorem of Singler [2] from 1938, which we state now. Theorem 2.2. (Singler) Let q be a power of a prime1 . The automorphism group of P G(2, q) contains a cyclic subgroup hσi which acts regularly on points and regularly on lines. Singler’s result allows us to prove the following important lemma. Lemma 2.1. Let a be an integer. If there exists an a × (a2 − a + 1) latin rectangle without any rainbow a×2 subrectangles, then a−1 ∈ P. If a−1 is a prime power, then there exists an a × (a2 − a + 1) latin rectangle without any rainbow a × 2 subrectangles. Proof. Suppose R is an a × (a2 − a + 1) latin rectangle without a rainbow a × 2 subrectangle. Note that any two columns of R have a common element. Given any column C of R, each of the other a(a − 1) columns must contain one of the a elements in C. Thus each of the elements of C is, on average, in at least a − 1 other columns. Now, assume that one of these elements is in strictly fewer than a − 1 other columns. Then another element would have to be in at least a other columns. Clearly this is impossible – along with C such an element is in at least a+1 columns, implying that two must be in a common row. Thus every element is in exactly a different columns and every column has exactly a different elements. Furthermore it is easy to see by the same reasoning that no two columns can intersect in more than one element. Thus the columns of R satisfy the axioms of a projective plane of order a − 1. We now assume that a − 1 is a prime power and consider a projective plane P of order a − 1. We aim to construct a latin rectangle of the asserted size. Take any line B from P and use its elements to form the first edge of our latin rectangle; the orders of the elements are not important. By Theorem 2.2 there exists a cyclic subgroup hσi which acts regularly on the points and regularly on lines. For each i, apply σ to the ith member of the first column to obtain the ith member of 1

In this paper, we consider 1 to be a prime power.

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the second column. In general, obtain each successive column by applying σ to the previous. Recall that a projective plane of order a − 1 has (a − 1)2 + (a − 1) + 1 = a2 − a + 1 points, so σ is simply a cyclic ordering of these. Thus, the ith row is simply a list of all a2 − a + 1 elements which follows this ordering and starts with the ith member of the first column. Given this, it’s clear that each row has distinct entries. Furthermore, since B has distinct members it is also clear that each successive column has distinct entries. Indeed, if two entries were the same then that implies that the two rows are completely the same, as each successive entry is obtained by repeatedly applying σ. In particular their two entries in B are the same. But this is a contradiction as B was a single line in P and hence has distinct members. This completes the proof. Example 2.1. For the seven-point plane, consider σ = (1, 3, 5, 7, 2, 4, 6) and B = {1, 2, 4}. Then a corresponding latin rectangle is 1 2 4

3 5 4 6 6 1

7 2 4 1 3 5 3 5 7

6 7 2

Theorem 2.3. Let a, b, m, n be positive integers satisfying a ≤ b, m ≤ n, m < b, and n ≤ (a2 − a + 1)(b − 1). If a − 1 is a prime power, then it is possible to properly edge-color Km,n to forbid the existence of any rainbow Ka,b subgraph. That is, if a − 1 is a prime power, then Km,n 6→R Ka,b . Proof. Suppose a − 1 is a prime power. By Lemma 2.1, there exists an a × (a2 − a + 1) latin rectangle that contains no rainbow a × 2 subrectangle. We prove that there exists an m × n latin rectangle containing no rainbow a × b or b × a subrectangle. Because m < b, there are no b × a subrectangles of any m × n latin rectangle, so it suffices to construct an m × n latin rectangle with no rainbow a × b subrectangles. Furthermore, it is easy to see that it suffices to construct such a rectangle for the case in which m = b − 1 and n = (a2 − a + 1)(b − 1). Let m = b − 1 and n = (a2 − a + 1)(b − 1). Let L be an m × n latin rectangle. We first partition L into m subrectangles A1 , A2 , . . . , Am , each of size m × (a2 − a + 1). By the Pigeonhole Principle, any a × b subrectangle of L must contain at least two columns from Ak for some k ∈ {1, 2, . . . , m}. In other words, any a × b subrectangle of L contains an m × 2 subrectangle of Ak for some k ∈ {1, 2, . . . , m}. We will fill L with symbols in such a manner so as to ensure that, for any k ∈ {1, 2, . . . , m}, there is no rainbow m × 2 subrectangle of Ak , which will then imply the desired result. We may fill L with symbols so that, for any distinct j, k ∈ {1, 2, . . . , m}, no symbol appears in both Aj and Ak . This will ensure that the choice of symbols in Aj does 4

not affect where we may choose to place symbols in Ak and vice versa. Therefore, it suffices to show that we may fill an m × (a2 − a + 1) latin rectangle with symbols so that any two columns have a symbol in common. To do so, we simply extend the a × (a2 − a + 1) latin rectangle that we assumed exists to an m × (a2 − a + 1) latin rectangle. We close with a final result on how to generate a sequence of latin rectangles to avoid larger and larger rainbow subrectangles. Theorem 2.4. Let m, n, a, b, r, s be positive integers. Let J be the m×n all-1s matrix, and let A be an m × n latin rectangle consisting of positive integers whose maximum entry is t. Let B be an r × s latin rectangle whose entries are nonnegative integers. Let   A + tB11 J A + tB12 J · · · A + tB1r J    A + tB21 J A + tB22 J · · · A + tB2r J  A′ = J ⊗ A + tB ⊗ J =  . .. .. .. ..   . . . . A + tBr1 J A + tBr2 J · · · A + tBrr J Then A′ is an rm × sn latin rectangle. If A has no rainbow a × b subrectangles, then A′ has no rainbow r(a − 1) + 1 × s(b − 1) + 1 subrectangles. In particular, if Km,n 6→R Ka,b then Krm,rn 6→R Kr(a−1)+1,r(b−1)+1 . Proof. We defined A′ to be a block matrix with rs blocks, each of the form A + tBij J. Because t is the largest entry in A and the entries of A and B are nonnegative integers, two blocks A+tBij J and A+tBkℓ J cannot have any common entries unless Bij = Bkℓ . Therefore, since A and B are latin, it is easy to see that A′ must also be latin. Let T be an r(a − 1) + 1 × s(b − 1) + 1 subrectangle of A′ . By the pigeonhole principle, there must be a rows of T and b columns of T which intersect in a single block of A′ , say A + tBuv J. The intersection of these rows and columns forms an a × b subrectangle R of A + tBuv J. Since A contains no rainbow a × b subrectangles, A + tBuv J cannot contain a rainbow a × b subrectangle. This implies that R is not rainbow. Since R is a subrectangle of T , T cannot be rainbow. As T was arbitrary, this shows that A′ contains no rainbow r(a − 1) + 1 × s(b − 1) + 1 subrectangles. Now, suppose Km,n 6→R Ka,b . Then we may let A be an m × n latin rectangle that contains no rainbow a × b subrectangles and contains no rainbow b × a subrectangles. Setting r = s in the first part of the theorem, we obtain an rm × rn latin rectangle A′ that contains no rainbow r(a − 1) + 1 × r(b − 1) + 1 subrectangles. However, since A also has no rainbow b×a subrectangles, we may interchange the roles of a and b in the first part of the theorem to see that A′ also has no rainbow r(b − 1) + 1 × r(a − 1) + 1 subrectangles. Hence, Krm,rn 6→R Kr(a−1)+1,r(b−1)+1 . 5

3

Anti-Ramsey Numbers

Returning to Theorem 2.3, we find it particularly interesting to consider the case a = 2, n = 2m = 2b. That is, we wish to find positive integers m such that any proper edge-coloring of Km,2m contains a rainbow K2,m subgraph. Theorem 2.1 shows that any proper edge-coloring of K2,4 contains a rainbow K2,2 subgraph, and the following theorem deals with the case m = 3. Theorem 3.1. Any properly edge-colored K3,6 contains a rainbow K2,3 subgraph. Proof. We prove the equivalent statement that every 3 × 6 latin rectangle contains a rainbow 2 × 3 or 3 × 2 subrectangle. To do so, suppose there exists some latin rectangle R with no 2 × 3 or 3 × 2 subrectangle. We will refer to the symbols in R as ”colors” in order to maintain the correspondence between R and a properly edge-colored K3,6 . We will let [r1 , r2 : c1 , c2 , c3 ] denote the 2 × 3 subrectangle of R that is the intersection of rows r1 and r2 and columns c1 , c2 , and c3 . Let us denote the color in the ith row and the j th column of R by Rij . Note that we may swap any two columns of R without changing the fact that R does not contain a rainbow 2 × 3 or 3 × 2 subrectangle. We will let S(i, j) denote the operation of swapping columns i and j of R. Furthermore, at any time, we may exchange any two colors r and s so that all entries of R colored r are recolored s and vice versa. Let C(r, s) denote the operation of exchanging colors r and s, and note that this operation does not change the fact that there is no rainbow 2 ×3 or 3 ×2 subrectangle of R. Even after swapping columns and exchanging colors of R, we will continue to refer to the rectangle as R. Now, call any coloring of R with the property that R1j = j for all j ∈ {1, 2, . . . , 6} a “primal” coloring. Without loss of generality, we may assume R is primally colored. Consider the 2 × 3 subrectangle [1,2:1,2,3] of R. Because this subrectangle is not rainbow, one or more of the following equalities must hold: R22 = 1, R23 = 1, R21 = 2, R23 = 2, R21 = 3, R22 = 3. Suppose R23 = 1. Then, performing the operation S(2, 3) followed by C(2, 3), we reach a primal coloring in which R22 = 1. Next, suppose that R21 = 2. Performing the operation S(1, 2) followed by C(1, 2), we reach a primal coloring in which R22 = 1. A similar argument shows that we may assume, without loss of generality, that R is primally colored and R22 = 1. Now, consider the 2 × 3 subrectangle [1, 2 : 3, 4, 5]. By the same argument as before, we see that, without loss of generality, we may assume that R is primally colored, R22 = 1, and R24 = 3. This is because, in order to ensure that R is primally colored with R24 = 3, we only need to use some combination of some of the operations S(3, 4), S(3, 5), S(4, 5), C(3, 4), C(3, 5), and C(4, 5), none of which change the fact that R22 = 1. Consider the subrectangle [1, 2 : 1, 3, 6] of 6

R. Because this subrectangle is not rainbow, we require either R21 = 6 or R23 = 6. If R23 = 6, perform the operations S(1, 3), S(2, 4), C(1, 3), and C(2, 4) to obtain a primal coloring of R in which R22 = 1, R24 = 3, and R21 = 6. If R21 = 6, then we do not need to perform any operations to obtain such a coloring. If we now consider the subrectangle [1, 2 : 1, 3, 5], it is easy to see that we must have R23 = 5. Considering [1, 2 : 2, 3, 6], we see that we must have R25 = 4. If we now consider [1, 3 : 3, 4, 5], we see that we must have R33 = 4, R34 = 5, or R35 = 6. No matter what, there must be some column A of R that contains the colors 3, 4, and 5. Similarly, if we consider [1, 3 : 1, 2, 6], we see that we must have R31 = 2, R32 = 6, or R36 = 1. No matter what, there must be some column B of R containing the colors 1, 2, and 6. However, this is a contradiction because the union of A and B is a rainbow 3 × 2 subrectangle of R. Given Ka,b we wish to find the “smallest” complete bipartite graph Km,n for which Km,n →R Ka,b . Here we use the number of vertices m + n as our “smallest” metric. To this end we define the following. Definition 3.1. Let Km,n be the complete bipartite graph with the fewest vertices for which Km,n →R Ka,b . Then we define the vertex anti-Ramsey number ARV (Ka,b ) to be the number of vertices m + n of this minimal example. We begin the study of these numbers by finding some specific values. Theorem 3.2. For any integer b ≥ 2, ARV (K2,b ) = 3b. Proof. By Theorem 2.1, K2,3b−2 →R K2,b , so ARV (K2,b ) ≤ 3b. Now, let m and n be positive integers with m + n < 3b. We wish to show that Km,n 6→R K2,b . We may assume 2 ≤ m ≤ n. If m < b, then we may use Theorem 2.3 (with a = 2) to deduce that Km,n 6→R K2,b . Therefore, let us assume m ≥ b. Since m + n < 3b, n < 2b. Since m ≤ n, it is possible to properly edge-color Km,n with n colors. Such a coloring must necessarily forbid the existence of a rainbow K2,b subgraph because any K2,b subgraph has 2b edges. Definition 3.2. Let Km,n be a complete bipartite graph with the fewest edges for which Km,n →R Ka,b . Then we define the edge anti-Ramsey number ARE (Ka,b ) to be the number of edges mn of this minimal example. Note that this generalizes the notion of size anti-Ramsey numbers given in [1]. From Theorem 3.2 we know that ARE (K2,3 ) ≤ 18. We initiate the study of these size anti-Ramsey numbers with the following theorem. 7

Theorem 3.3. If a and b are integers such that a−1 is a prime power and b ≥ a(a−1), then ARE (Ka,b ) = a2 (a − 1)(b − 1) + ab. Proof. Suppose a and b are integers with a − 1 a prime power and b ≥ a(a − 1). By Theorem 2.1, Ka,(a2 −a+1)(b−1)+1 →R Ka,b , so ARE (K2,b ) ≤ a((a2 − a + 1)(b − 1) + 1) = a2 (a − 1)(b − 1) + ab. Now, let m and n be positive integers with mn < a2 (a − 1)(b − 1) + ab. We wish to show that Km,n 6→R Ka,b . We may assume a ≤ m ≤ n. Since m ≤ n, it is possible to properly edge-color Km,n with n colors. If n < ab, then such a coloring must necessarily forbid the existence of a rainbow Ka,b subgraph because any Ka,b subgraph has ab edges. Therefore, let us assume that n ≥ ab. Since mn < a2 (a − 1)(b − 1) + ab and a(a − 1) ≤ b, m